Solve ∫ (from - 4 to - 1) of (2x^2+10x+8) wrt x

$$\int _ { - 4 } ^ { - 1 } ( 2 x ^ { 2 } + 10 x + 8 ) d x$$

Evaluate

$$-9$$

Solution Steps

Evaluate the indefinite integral first.

$$\int 2x^{2}+10x+8\mathrm{d}x$$

Integrate the sum term by term.

$$\int 2x^{2}\mathrm{d}x+\int 10x\mathrm{d}x+\int 8\mathrm{d}x$$

Factor out the constant in each of the terms.

$$2\int x^{2}\mathrm{d}x+10\int x\mathrm{d}x+\int 8\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{2}\mathrm{d}x with \frac{x^{3}}{3}. Multiply 2 times \frac{x^{3}}{3}.

$$\frac{2x^{3}}{3}+10\int x\mathrm{d}x+\int 8\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x\mathrm{d}x with \frac{x^{2}}{2}. Multiply 10 times \frac{x^{2}}{2}.

$$\frac{2x^{3}}{3}+5x^{2}+\int 8\mathrm{d}x$$

Find the integral of 8 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{2x^{3}}{3}+5x^{2}+8x$$

The definite integral is the antiderivative of the expression evaluated at the upper limit of integration minus the antiderivative evaluated at the lower limit of integration.

$$\frac{2}{3}\left(-1\right)^{3}+5\left(-1\right)^{2}+8\left(-1\right)-\left(\frac{2}{3}\left(-4\right)^{3}+5\left(-4\right)^{2}+8\left(-4\right)\right)$$

Simplify.

$$-9$$